SATHEE: Area Question 4

Area Question 4

Question 4

The area (in sq units) of the region $A = (x, y) : \frac{y^{2}}{2} \leq x \leq y + 4$ is

(2019 Main, 9 April II) (a) 30 (b) $\frac{53}{3}$ (c) 16 (d) 18

Show Answer

Solution:

Given region $A = (x, y) : \frac{y^{2}}{2} \leq x \leq y + 4$

$∴ \frac{y^{2}}{2} = x$

$\Rightarrow y^{2} = 2 x$

and $x = y + 4 \Rightarrow y = x - 4$

Graphical representation of $A$ is

On substituting $y = x - 4$ from Eq. (ii) to Eq. (i), we get

$\begin{aligned} (x - 4)^{2} & = 2 x \Rightarrow & x^{2} - 8 x + 16 & = 2 x \Rightarrow & x^{2} - 10 x + 16 & = 0 \Rightarrow & (x - 2) (x - 8) & = 0 \Rightarrow & x & = 2, 8 ∴ & y & = - 2, 4 \end{aligned}$

[from Eq. (ii)]

So, the point of intersection of Eqs. (i) and

(ii) are $P (2, - 2)$ and $Q (8, 4)$ .

Now, the area enclosed by the region $A$

$\begin{aligned} = \int_{- 2}^{4} (y + 4) - \frac{y^{2}}{2} d y = \frac{y^{2}}{2} + 4 y - {\frac{y^{3}}{6}}_{- 2}^{4} & = \frac{16}{2} + 16 - \frac{64}{6} - \frac{4}{2} - 8 + \frac{8}{6} & = 8 + 16 - \frac{32}{3} - 2 + 8 - \frac{4}{3} & = 30 - 12 = 18 sq unit. \end{aligned}$

Area Question 4

Question 4

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Area Question 4

Question 4

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu